Question: Simplify and expand the following expression: $ \dfrac{2t - 9}{3t - 5}-\dfrac{5t + 1}{5t - 5} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3t - 5)(5t - 5)$ Multiply the first term by $\dfrac{5t - 5}{5t - 5}$ $ \begin{align*} \dfrac{2t - 9}{3t - 5} \times \dfrac{5t - 5}{5t - 5} & = \dfrac{(2t - 9)(5t - 5)}{(3t - 5)(5t - 5)} \\ & = \dfrac{10t^2 - 55t + 45}{(3t - 5)(5t - 5)}\end{align*} $ Multiply the second term by $\dfrac{3t - 5}{3t - 5}$ $ \begin{align*} \dfrac{5t + 1}{5t - 5} \times \dfrac{3t - 5}{3t - 5} & = \dfrac{(5t + 1)(3t - 5)}{(5t - 5)(3t - 5)} \\ & = \dfrac{15t^2 - 22t - 5}{(5t - 5)(3t - 5)}\end{align*} $ Now we have: $ = \dfrac{10t^2 - 55t + 45}{(3t - 5)(5t - 5)} - \dfrac{15t^2 - 22t - 5}{(5t - 5)(3t - 5)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{10t^2 - 55t + 45 - (15t^2 - 22t - 5)}{(3t - 5)(5t - 5)} $ $ = \dfrac{10t^2 - 55t + 45 - 15t^2 + 22t + 5}{(3t - 5)(5t - 5)} $ $ = \dfrac{-5t^2 - 33t + 50}{(3t - 5)(5t - 5)}$ Expand the denominator: $ = \dfrac{-5t^2 - 33t + 50}{15t^2 - 40t + 25}$